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3 years ago

A man pulls a 50.0 kg box with a rope parallel to the ground he pulls the box 10.0 m about how much work has he done

2 answers:

4 0

Your answer is 5000 J

when W(work) = F X when F= the force and X= the displacment

and F(g) = M a(g) when M= mass and a = the acceleration and in our question
, the force is the gravitational force and a= 9.8 m/S2 we can assume as 10 m/s2

and when we have M= 50 Kg
so by substitution:
F= 50 x 10 = 500 N

and by substitution in work equation: when x = 10 m
∴ W = 500 x 10 = 5000 j

Mashutka [201]3 years ago

4 0

The weight of the box is (using an approximate value $g=10m/s^2$ :
$W=mg=50Kg\cdot10m/s^2=500 N$
Therefore, the man must apply an equal force to pull the box at a height of h=10.0m. The work he has done is equal to the product between this force and the displacement, which is the height h:
$W=Fd=500N \cdot 10m=5000 J$

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3 years ago

A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th

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Assume that $\rm g= 9.81\; N\cdot kg^{-1}$ ; $\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}$ .

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Diameter: $d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m$ .
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$V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}$ .

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$\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}$ .

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$W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N$ .

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately $\rm 0.0114009\; m^{3}$ of water. The buoyant force on the disk will be:

$\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}$ .

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

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4 years ago

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The correct option would be B. volts

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